# Normal matrices and the SVD

I've been stuck in this problem on SVD.

Let $A\in M_n(\mathbb{C})$, with $rank(A)=k$. Prove that A is normal if and only if exists an orthonormal set in $\mathbb{C}^n$, $\lbrace z_1,...,z_k \rbrace$, such that $|z_i^{*}Az_i|=\sigma_i$ for all $i=1,2,3,...,k$. With $\sigma_i$ singular values of A.

My attempt: for $\Leftarrow|)$ i've seen that if A has $rank(k)$ then exists $U,V$ unitary matrices such that: $A=U \Sigma V^*$, with $\Sigma$ a rectangular matrix containing the $k$ singular values. Since the singular values can be written as $|z_i^{*}Az_i|=\sigma_i$ then when you transpose $\Sigma$, the elements on the diagonal are preserved. Then I've got for instance:

$AA^* =U\Sigma V^* V\Sigma^* U^*=U\Sigma\Sigma^*U^*$. And

$A^*A=V\Sigma^*\Sigma V^*$. From here... can I conclude thar A is normal?

Also the other implication is bothering me a lot, so any suggestion would be appreciated. Thanks!

My instructor tells me that i should use the fact that certain norm i the sum of the squares of the singular values of the matrix,

• The $\implies$ implication is easy using the spectral theorem Nov 22 '16 at 6:16
• Ohh really? I didn't try it. Let me dig on that idea. Nov 22 '16 at 6:17
• First thought for the $\Leftarrow$ part: note that $A$ has a Schur triangularization so that $A$ is upper triangular without loss of generality. Then, show that the conditions on the singular values imply that $A$ must be diagonal. Nov 22 '16 at 6:25
• It wouldn't be just similar to an upper triangular? instead of being itself upper triangular? Though, it would have the eigenvalues on the diagonal, and the same spectre., no? Nov 22 '16 at 6:31
• The upper triangular matrix is unitarily similar, so it has the same eigenvalues and singular values. Nov 22 '16 at 13:01

You have solved part of the problem. Just carry through with the $\Sigma$ gymnastics.
$\Leftarrow$
\begin{align} \mathbf{A} &= \mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\ % &= % U \left[ \begin{array}{cc} \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}} \end{array} \right] % Sigma \left[ \begin{array}{cccccc} \sigma_{1} & 0 & \dots & & & \dots & 0 \\ 0 & \sigma_{2} \\ \vdots && \ddots \\ & & & \sigma_{k} \\ & & & & 0 & \\ \vdots &&&&&\ddots \\ 0 & & & & & & 0 \\ \end{array} \right] % V \left[ \begin{array}{c} \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\ \color{red}{\mathbf{V}_{\mathcal{N}}}^{*} \end{array} \right] \\ % & = % U \left[ \begin{array}{cccccccc} \color{blue}{u_{1}} & \dots & \color{blue}{u_{k}} & \color{red}{u_{k+1}} & \dots & \color{red}{u_{n}} \end{array} \right] % Sigma \left[ \begin{array}{cc} \mathbf{S}_{k\times k} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{array} \right] % V \left[ \begin{array}{c} \color{blue}{v_{1}^{*}} \\ \vdots \\ \color{blue}{v_{k}^{*}} \\ \color{red}{v_{k+1}^{*}} \\ \vdots \\ \color{red}{v_{n}^{*}} \end{array} \right] % \end{align} Because the number of rows is the same as the number of columns, $$\Sigma \, \Sigma^{\mathrm{T}} = \Sigma ^{\mathrm{T}} \Sigma = \Sigma^{2} = \left[ \begin{array}{cc} \mathbf{S}^{2} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{array} \right]$$ The matrix product $$\mathbf{A}^{*} \mathbf{A} = \mathbf{V} \, \Sigma^{\mathrm{T}} \, \Sigma \mathbf{V}^{*}.$$ Using the thin SVD (ignoring the nullspace terms) we can write $$\mathbf{A}^{*} \mathbf{A} = \color{blue}{\mathbf{V}_{\mathcal{R}}} \, \mathbf{S}^{2} \, \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*}.$$ The transpose is $$\left( \mathbf{A}^{*} \mathbf{A} \right)^{*} = % \left( \color{blue}{\mathbf{V}_{\mathcal{R}}} \, \mathbf{S}^{2} \, \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \right)^{*} = % \left( \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \right)^{*} \, \left( \mathbf{S}^{2} \right)^{\mathrm{T}} \, \left( \color{blue}{\mathbf{V}_{\mathcal{R}}} \right)^{*} = % \color{blue}{\mathbf{V}_{\mathcal{R}}} \, \mathbf{S}^{2} \, \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*}.$$ Therefore $$\left[ \mathbf{A}^{*} \mathbf{A} \right] = % \mathbf{A}^{*} \mathbf{A} - \mathbf{A} \, \mathbf{A}^{*} = % \mathbf{0}.$$